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Case (3): If s = 3 then equality (10) becomes
n
1 1
(3) = + 3 (1=n); (21)
k3 2
k=1

where
âˆž
eâˆ’u=x 2
3 (x) = uu du
e âˆ’1
0

whose asymptotic expansion is
âˆž
Bk (k + 1)xk+2 :
3 (x) =
k=0

Computation of [p=p] : Let us deÃ¿ne the derivative of B by
2
3 (x)=x

âˆ’B ; xk := B; kxkâˆ’1 = kBkâˆ’1 ; kÂ¿1;

âˆ’B ; 1 := 0:
So, the functional B admits an integral representation:
+iâˆž
cos( x)
2
k
xk
B ;x =i d x; âˆ’1 Â¡ Â¡ 0:
sin3 ( x)
âˆ’iâˆž

Let ( n )n be the sequence of orthogonal polynomial with respect to the sequence
âˆ’B0 :=0; âˆ’B1 = B0 ; âˆ’B2 = 2B1 ; âˆ’B3 = 3B2 ; : : : :
The linear form B is not deÃ¿nite and so the polynomials n are not of exact degree n.
More precisely, 2n has degree 2n and 2n+1 = 2n . For the general theory of orthogonal poly-
nomials with respect to a nondeÃ¿nite functional, the reader is referred to Draux [15]. If we take
= âˆ’ 1 , the weight cos x=sin3 ( x) d x on the line âˆ’ 1 + iR becomes sinh t=cosh3 t dt on R, which
2 2
is symmetrical around 0. So, 2n (it âˆ’ 1 ) only contains even power of t and we can write
2
1 2
2n (it âˆ’ 2 ) = Wn (t ), Wn of exact degree n. Thus Wn satisÃ¿es

t sinh t
Wn (t 2 )Wm (t 2 ) dt = 0; n = m:
cosh3 t
R

The weight t sinh t=cosh3 t equals (1=4 3 )| ( 1 + it)|8 | (2it)|2 and has been studied by Wilson
2
[33,3]:
n n+k y+k y
n
nÂ¿0; 2n (y) = : (22)
k k k k
k=0
244 M. PrÃƒ vost / Journal of Computational and Applied Mathematics 122 (2000) 231â€“250
e

Let the polynomial associated to 2n :
2n

2n (x)
âˆ’ 2n (t)
2n (t) = âˆ’B ; ; B acts on x:
xâˆ’t
For the computation of 2n , we need to expand the polynomial
x+k x t+k t
âˆ’
k k k k
:
xâˆ’t
On the Newton basis with the abscissa {0; 1; âˆ’1; : : : ; n; âˆ’n}
x+k x t+k t
âˆ’ 2k
k k k k N2k (t) Niâˆ’1 (x)
= ;
xâˆ’t Ni (t) [(i + 1)=2]
i=1

where N0 (x):=1, N1 (x) = 1 , N2 (x) = 1 x+1 ; : : : ; N2i (x) =
x x x x+i x x+i
N2i+1 (x) = .
1 i i i+1 i
By recurrence, the values âˆ’B ; Ni (x) can be found in
(âˆ’1)i
i âˆˆ N; âˆ’B ; N2i (x) = 0; âˆ’B ; N2i+1 (x) = :
(i + 1)2
Using the linearity of B , we get the expression of 2n :

t+k tâˆ’i
n n+k
n k
k âˆ’i k âˆ’i
(âˆ’1)i+1
2n (t) = âˆˆ P2nâˆ’2 : (23)
2
i3
k k k
i=1
k=0

i
Eq. (16) implies that
d3 2n (t) âˆˆ N; âˆ€t âˆˆ N:
n

The link between 2n , and the ApÃƒ ryâ€™s numbers an , bn is given by taking y = n in (22) and
e
2n
t = n in (23):
2 2
n n+k
n

2n (n) = = bn ;
k k
k=0

n
1 1
2n (n) + 2n (n) = an :
k3 2
k=1

ApÃƒ ry was the Ã¿rst to prove irrationality of (3). He only used recurrence relation between the an
e
and bn . We end the proof of irrationality of (3) with the error term for the PadÃƒ approximation.
e
Let us recall equality (21),
n
1 1 1
(3) = + 3
k3 2 n
k=1
M. PrÃƒ vost / Journal of Computational and Applied Mathematics 122 (2000) 231â€“250
e 245

in which we replace the unknown term 3 (1=n) by its PadÃƒ approximant [2n=2n] 3 (x = 1=n). It arises
e
the following approximation for (3):
n
1 1 2n (n)
(3) â‰ˆ +
k3 2 2n (n)
k=1

and the expression
n
1
2d3 d3
en = 2n (n) (3) âˆ’ 2 2n (n) + 2n (n)
n n
k3
k=1

and d3
will be a Diophantine approximation, if we prove that limn en = 0 (since 2n (n) 2n (n) are
n
integer).
Let us estimate the error en . The method is the same as for (2):
âˆ’1
2n (t )
2
3 (t) âˆ’ [2n=2n] 3 (t) = 3 (t) âˆ’ t [2n âˆ’ 2=2n] 2 (t) = 3 (t) âˆ’ :
3 =t
2n (t
âˆ’1 )

The integral representation of B gives
t 2i +iâˆž 2
2n (x)
cos x
3 (t) âˆ’ [2n=2n] 3 (t) = âˆ’ 2 d x:
1 âˆ’ xt sin3 x
2n (t
âˆ’1 )
âˆ’iâˆž

The previous expression implies that the error 3 (t) âˆ’ [2n=2n] 3 (t) is nonzero, and also that
2
t 1 u sinh u
Wn2 (u2 ) t âˆˆ R+ :
| 3 (t) âˆ’ [2n=2n] 3 (t)|6 Â· Â· du;
cosh3 u
2
2n (t
âˆ’1 ) 1 + t=2
R

From the expression of the integral (see [33]) we get
42
| 3 (1=n) âˆ’ [2n=2n] 3 (1=n)|6 :
2
(2n + 1)2 2n (n)

The error term in the PadÃƒ approximation satisÃ¿es
e
n
42
1
2 (3) âˆ’ 2 âˆ’ [2n=2n] 3 (1=n) 6 2
k3 (2n + 1)2 2n (n)
k=1

and the error term en satisÃ¿es
n
82 d3
1 n
2d3 d3
|en | = 2n (n) (3) âˆ’ 2 2n (n) + 2n (n) 6 :
n n
k3 (2n + 1)2 2n (n)
k=1
âˆš
2)4n nâˆ’3=2 [14], and so we get, since dn = en(1+o(1)) ,
2n (n) = bn implies that 2n (n) = A(1 +
|2d3 bn (3) âˆ’ 2d3 an | â†’ 0;
n n
(24)
n â†’ âˆž;
where 2d3 bn and 2d3 an are integers.
n n
The above relation (24) shows that (3) is irrational.
246 M. PrÃƒ vost / Journal of Computational and Applied Mathematics 122 (2000) 231â€“250
e

Of course, using the connection between PadÃƒ approximation and -algorithm, the Diophantine
e
approximation of (3) can be constructed by means of the following -array: an =bn = n 1=k 3 +
k=1
(0) (0) n 3
4n (Tm ) = 4n ( k=1 1=k + Tm ); where Tm is the partial sum of the asymptotic series (nonconvergent)
m
1
Tm = 2 k=1 Bk (k + 1)1=nk :
We get the following -arrays for n = 1,
ï£® ï£¹
0
ï£¯ ï£º
ï£¯0 0 ï£º
ï£¯ ï£º
ï£¯ ï£º
ï£¯ (0) ï£º (0)
1+ 1 âˆ— = 6 = a1 =b1
4 ï£º; (Apery s numbers);
ï£¯ 1 1=2 2=5 = 4
ï£¯ ï£º 2 5
ï£¯ ï£º
ï£¯ 0 1=3 ï£º
ï£° ï£»
1=2
and for n = 2,
ï£® ï£¹
0
ï£¯ ï£º
ï£¯0 ï£º
0
ï£¯ ï£º
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